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Ch. 5 - Integrals
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 5, Problem 5.PE.25

Find the areas of the regions enclosed by the curves and lines in Exercises 15–26.


y = 2 sin x, y = sin 2x, 0 ≤ x ≤ π

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First, identify the curves given: \(y = 2 \sin x\) and \(y = \sin 2x\), and the interval for \(x\) is \(0 \leq x \leq \pi\).
Next, find the points of intersection between the two curves by setting \(2 \sin x = \sin 2x\) and solving for \(x\) within the interval \([0, \pi]\).
Recall that \(\sin 2x = 2 \sin x \cos x\), so the equation becomes \(2 \sin x = 2 \sin x \cos x\). Simplify this to find the values of \(x\) where the curves intersect.
Determine which curve is on top (greater \(y\)-value) between each pair of intersection points by testing values of \(x\) in the intervals formed by the intersection points.
Set up the integral(s) for the area(s) between the curves using the formula \(\text{Area} = \int_a^b \left( \text{top curve} - \text{bottom curve} \right) \, dx\), where \(a\) and \(b\) are the intersection points or the interval endpoints.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Finding Area Between Curves

To find the area enclosed between two curves, subtract the lower function from the upper function over the given interval and integrate the difference. This gives the net area between the curves, accounting for where one function lies above the other.
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Properties of Sine Functions

Understanding the behavior of sine functions, such as y = 2 sin x and y = sin 2x, is essential. Knowing their periodicity, amplitude, and zeros helps determine which curve is on top and the points of intersection within the interval 0 ≤ x ≤ π.
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Definite Integration over a Specified Interval

Definite integration calculates the exact area under a curve between two limits. Here, integrating from 0 to π allows finding the total enclosed area between the given sine curves within the specified domain.
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